Standardization of a random variable with normal distribution

Let $$X \sim N(\mu, \sigma^2)$$ \begin{align*} P\bigg(\frac{X - \mu}{\sigma} \leqslant x\bigg) &= P(X \leqslant x\cdot\sigma + \mu)\\ &= \int_{-\infty}^{x\cdot\sigma + \mu}{\frac{1}{\sqrt{2\pi}\sigma}\cdot\exp\bigg(-\frac{(x-\mu)^2}{2\sigma^2}\bigg)\,dx}\\ &= \int_{-\infty}^{x\cdot\sigma + \mu}{\frac{1}{\sqrt{2\pi}\sigma}\cdot\exp\Bigg(-\frac{1}{2}\cdot\bigg(\frac{x-\mu}{\sigma}\bigg)^2\Bigg)\,dx} \end{align*} We substitute with $$\frac{x-\mu}{\sigma} = y \quad\Rightarrow\quad \frac{1}{\sigma}\,dx = dy \quad\Rightarrow\quad dx = \sigma \,dy$$ so: \begin{align*} &= \int_{-\infty}^y {\frac{1}{\sqrt{2\pi}\sigma}\cdot\exp\bigg(-\frac{1}{2}\cdot y^2\bigg)\cdot\sigma \, dy}\\ &= \int_{-\infty}^{y}{\frac{1}{\sqrt{2\pi}}\cdot\exp\bigg(-\frac{y^2}{2}\bigg) \,dy}\\ &= \Phi(y) \end{align*} Therefore $$\frac{X - \mu}{\sigma} \sim N(0, 1)$$ Is that correct?

• Yes ${}{}{}{}{}$ Commented Dec 4, 2014 at 21:58

But I wouldn't go from $\dfrac{dx}\sigma=dy$ to $dx = \sigma\,dy$. Instead, since the integral is $$\int_{-\infty}^{\mu+x\sigma} \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}2\left(\frac{x-\mu}{\sigma}\right)^2\right) \frac{dx}\sigma,$$ I'd just put $dy$ in place of $\dfrac{dx}\sigma$.